Printable · GCSE Foundation · ages 14-16
Probability worksheet — GCSE Foundation
Fifteen questions across the probability statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Probability worksheet — GCSE Foundation
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- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (a) 0.36 — Relative frequency is the number of times the event happened divided by the total number of trials: 18 ÷ 50 = 0.36. Dividing by 100 instead of the actual 50 spins gives 18 ÷ 100 = 0.18. Finding the relative frequency of NOT landing on green, using 50 − 18 = 32 spins, gives 32 ÷ 50 = 0.64. Misplacing the decimal point in the division, so that 18 ÷ 50 is carried out as 18 ÷ 500, gives 0.036 — a tenth of the correct value.
- (c) 3 — There are 3 outcomes for the first draw and 3 for the second, giving 3 × 3 = 9 ordered pairs in total: first draw 1 with second draw 1, 2 or 3; first draw 2 with second draw 1, 2 or 3; and first draw 3 with second draw 1, 2 or 3. Checking the list, the only pairs with both numbers the same are 1 with 1, 2 with 2, and 3 with 3, so there are 3. A candidate who answers 9 has counted every outcome instead of only the matching ones. A candidate who answers 6 has mistakenly counted pairs such as 1 with 2 and 2 with 1 as matching because they contain the same two digits. A candidate who answers 1 has stopped after finding only the first matching pair in the list.
- (c) The elements that are in A but not in B — The symbol ∩ means 'and', so A ∩ B′ means 'in A and also in the complement of B'. B′ means 'not in B'. So A ∩ B′ describes everything that is in A but not in B. The elements in both A and B describes A ∩ B, without the dash on B. The elements in B but not in A describes B ∩ A′, with the dash on A instead of B. The elements in neither A nor B describes (A ∪ B)′, the region outside both circles entirely.
- (d) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
- (a) 16 — Method: keep the two rolls in order, first roll then second roll, and count every ordered pair the dice can land on, so a (2, 3) result is counted separately from a (3, 2) result. Working: the first roll can land on any of 4 numbers, and for each of those the second roll can also land on any of 4 numbers, giving 4 × 4 = 16 ordered pairs. Answer: 16. Watch out: writing down 8 comes from adding the two rolls' counts, 4 + 4, instead of multiplying them. Writing down 10 comes from listing unordered pairs, so (2, 3) and (3, 2) are counted as the same entry — that gives the 4 doubles plus 6 mixed pairs, 10 in total, instead of 16 ordered pairs. And writing down 4 comes from counting only one roll's outcomes and never pairing the two rolls together at all.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (a) 17/32 — Because the counter is replaced, each pick is independent with P(red) = 5/8 and P(blue) = 3/8 every time. Both red has probability 5/8 × 5/8 = 25/64, and both blue has probability 3/8 × 3/8 = 9/64. 'Same colour' means either of these, so add them: 25/64 + 9/64 = 34/64 = 17/32. Giving 25/64 finds only the probability of both counters being red, leaving out both counters being blue, which also counts as the same colour. Giving 15/64 finds the probability of one red and one blue counter in just one of the two possible orders — the opposite of what was asked, and only half of it. Giving 13/28 works out the probabilities as if the counter had NOT been replaced, using 5/8 × 4/7 and 3/8 × 2/7, even though the question states it was put back.
- (c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
- (a) 90% — Method: turn the fraction into a decimal by dividing the numerator by the denominator, then multiply the decimal by 100 to turn it into a percentage. Working: 9 ÷ 10 = 0.9, and 0.9 × 100 = 90. Answer: 90%, which is the same probability written in a third form, and all three forms sit at the same point on the 0 to 1 scale. The distractors: 9% comes from reading the numerator of the fraction straight off as the percentage; 10% comes from reading the denominator straight off as the percentage; 0.9% comes from correctly finding the decimal 0.9 and then writing a percentage sign after it without multiplying by 100.
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